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Geometry Difficulty 7.8 National Olympiad, round 2 Prove it Baltic Way

Let $ABC$ be a triangle with circumcircle $\Gamma$ and circumcenter $O$ . Denote by $M$ the midpoint of $BC$ . Point $D$ is the reflection of $A$ over $BC$ , and $E$ is the intersection of $\Gamma$ and ray $MD$ . Let $S$ be the circumcenter of triangle $ADE$ . Prove that $A, E, M, O$ , and $S$ are concyclic.

Solution

Solution. Take $S$ to be the point such that $CPSQ$ is a parallelogram, as seen in figure 20. For points $X, Y, Z$ let $\text{rot } XYZ$ denote the morphism on translations induced by the rotation that takes line $XY$ to line $XZ$ , modulo half turn. As $ABCD$ is a cyclic quadrilateral it follows that $\text{rot } BAD = \text{rot } BCD$ . It follows that $\text{rot } PAD = \text{rot } BCP$ and hence $\text{rot } PAQ = \text{rot } QCP = \text{rot } PSQ$ . It follows that $S$ lies on $\Gamma$ .
Using the cyclic quadrilateral AQRS it follows that $\text{rot } RSQ = \text{rot } RAQ$ . Considering the cyclic quadrilateral ABCD it follows that $\text{rot } RAQ = \text{rot } RAD = \text{rot } RCD = \text{rot } RCP$ . Hence, $\text{rot } RSQ = \text{rot } RCP$ . As lines CP and SQ are parallel it follows that line SR is parallel to line CR. Now R is a common point so it follows that line SR = CR = CS.
As lines CS and PQ are diagonals in a parallelogram CPSQ it follows that line CR = CS passes through M, the midpoint of linesegment PQ.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.